# Question #6de4a

##### 2 Answers

The locus of all such points is a circle of radius

(3)

#### Explanation:

The circle with equation

All chords sub-tending a *right* angle in this circle (as all chords sub-tending a fixed angle in any circle) have fixed length.

A triangle formed by a chord sub-tending the *right* angle and two radii to its ends is a *right isosceles* triangle with a chord as a hypotenuse and each radius as a cathetus.

Therefore, the length of any chord, sub-tending the *right* angle in a circle of a radius

The segment from the center of a circle to a midpoint of a triangle described above is an altitude, median and right angle bisector of this *right isosceles* triangle. The length of this segment can also be calculated using Pythagorean Theorem as

This length *right* angle in our circle.

Therefore, the locus of all such points is a circle of radius

(3)

**The Right Choice is Option**

#### Explanation:

Observe that,

**Origin**

The parametric eqns. of this circle are,

Let

of this circle.

Then,

the **X-axis.**

Since, **right angle** at the **Origin**

that

**X-axis.**

If

then, the mid-point

To find the **Locus** of M, let us set,

Eliminating

the general eqn. of the **desired locus** is

In case,

Hence, the right choice is option

**Enjoy Maths.!**